Sunday, January 3, 2010

Problem 412: Triangle, Angles, Congruence

Proposed Problem
Click the figure below to see the complete problem 412 about Triangle, Angles, Congruence.

Problem 412: Triangle, Angles, Congruence.
See also:
Complete Problem 412
Level: High School, SAT Prep, College geometry

6 comments:

  1. Another solution with full detail:
    http://img9.imageshack.us/img9/8220/problem412.png

    From C draw CE//AB and CE=AB ( see attached picture)
    ABEC is a parallelogram and ∠ (ECy)= 2x
    Triangle DEC is isosceles => ∠ (DEC)=1/2∠ (ECy)=x
    Quadrilateral BDCE is cyclic with 2 opposite sides paralleled
    So BDCE is an isosceles trapezoid
    And ∠ (BDA)= ∠ (ECy)=2x => Triangles ABD is isosceles
    In triangles ABD 4x=180-40 => x=35

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  2. My solution:
    Draw circle 0 (circumcircle of triangle ABD)
    Let E be the intersection of circle 0 with BC.
    Construct AE and DE
    Triangle BED is isosceles since ∠EBD =∠EAD = x
    So ∠BAE = ∠BDE = x
    Now ∠BDC =40º+2x so ∠EDC=40º+x since ∠BDE= x
    Triangles ABE and EDC are similar by SAS
    So ∠ DCE= x , finally x=35º since x+x+40º+2x=180º => 4x=140º => x=35º

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  3. Let the angle bisector of A meet BC at E and let BE = p

    Now < DAE = < DBE = x so ABED is cyclic

    So (a-p)a = bc......(1)
    Angie bisector theorem
    p/(a-p) = c/b .......(2)

    (1) X (2) gives us

    ap = c^2 hence BC is tangential to Tr ACE at A and so < C = x

    Now we know all the angles of Tr. ABC in terms of x and adding 4x+ 40 = 180 and x = 35

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Or far more simply because ABED is cyclic BE = DE and Tr.s ABE and CDE are congruent SAS and so < C = x and the result follows

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  5. Problem 412
    Brings BE=//DC then BECD is parallelogram so <BcE=x=< BAE=<BEA.Then BECA isisosceles trapezoid so <ACB=<AEB=x . Then 4x+40=180 so x=35.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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